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More on Kepler’s Laws
Kepler’s 3rd Law
• Can be shown that this also applies to an elliptical orbit with
replacement of r with a, where a is the semimajor axis.
2

 3
4

2
3
T 
a

K
a

S
 GMSun 
• Ks is independent of the planet mass, & is valid for any planet
• Note: If an object is orbiting another object, the value of K
will depend on the mass of the object being orbited.
• For example, for the Moon’s orbit around the Earth, KSun is
replaced with KEarth, where KEarth is obtained by replacing
MSun by MEarth in the above equation.
Kepler’s 3rd Law
The square of a
planet’s orbital period
is proportional to the
cube of its mean
distance from the Sun.
Solar System Data
Table 13-2, p. 370
Kepler’s Laws
can be derived from
Newton’s Laws.
In particular, Kepler’s 3rd Law follows directly from the
Universal Law Gravitation:
Equating the gravitational force with the centripetal force shows that,
for any two planets (assuming circular orbits, and that the only
gravitational influence is the Sun):
Irregularities in planetary motion led to the discovery of
Neptune, and irregularities in stellar motion have led to
the discovery of many planets outside our solar system.
Example 6-8: Where is Mars?
Mars’ period (its “year”) was first noted by Kepler to
be about 687 days (Earth-days), which is (687 d/365 d) =
1.88 yr (Earth years). Determine the mean distance of
Mars from the Sun using the Earth as a reference.
Example 6-9: The Sun’s mass determined.
Determine the mass of the Sun given the Earth’s
distance from the Sun as rES = 1.5  1011 m
“Weighing” the Sun!
• We’ve “weighed” the Earth, now lets “weigh” the Sun!!
Assume: Earth & Sun are perfect uniform spheres. & Earth orbit
is a perfect circle.
• Note: For Earth, Mass ME = 5.99  1024kg
Orbit period is T = 1 yr  3 107 s
Orbit radius r = 1.5  1011 m
So, orbit velocity is v = (2πr/T), v  3 104 m/s
• Gravitational Force between Earth & Sun: Fg = G[(MSME)/r2]
Circular orbit is circular  centripetal acceleration
Newton’s 2nd Law gives: ∑F = Fg = MEa = MEac = ME(v2)/r
OR: G[(MSME)/r2] = ME(v2)/r. If the Sun mass is unknown,
solve for it: MS = (v2r)/G  2  1030 kg  3.3  105 ME
Example 6-10: Lagrange point.
The mathematician Joseph-Louis Lagrange discovered five
special points in the vicinity of the Earth’s orbit about the
Sun where a small satellite (mass m) can orbit the Sun with
the same period T as Earth’s (= 1 year).
One of these “Lagrange Points,” called
L1, lies between the Earth (mass mE)
and Sun (mass mE), on the line
connecting them. That is, the Earth
and the satellite are always separated
by a distance d. If the Earth’s orbital
radius is RES, then the satellite’s
orbital radius is (RES - d). Determine d.
Sect. 6-6: Gravitational Field
The Gravitational Field is defined as
the gravitational force per unit mass:
The Gravitational Field due to a single
mass M is given by:
• A Gravitational Field  g, exists at all points in space.
• If a particle of mass m is placed at a point where the
gravitational field is g, it experiences a force:
• The field exerts a force F on the particle.
• The Gravitational Field g is defined as
• Gravitational Field = Gravitational Force experienced by a
“test” particle placed at that point divided by the mass of the test
particle.
– The presence of the test particle is not necessary for the field to exist
• A source particle creates the field
• The gravitational field g
vectors point in the direction
of the acceleration a particle
would experience if it were
placed in that field. Figure
• The field magnitude is that of
the freefall acceleration, g, at
that location.
• The gravitational field g
describes the “effect” that any
object has on the empty space
around itself in terms of the
force that would be present if
a second object were
somewhere in that space